What is the difference between printf() and puts() in C?

Question

puts is simpler than printf but be aware that the former automatically appends a newline. If that's not what you want, you can fputs your string to stdout or use printf.

(This is pointed out in a comment by Zan Lynx, but I think it deserves an aswer - given that the accepted answer doesn't mention it).

The essential difference between puts(mystr); and printf(mystr); is that in the latter the argument is interpreted as a formatting string. The result will be often the same (except for the added newline) if the string doesn't contain any control characters (%) but if you cannot rely on that (if mystr is a variable instead of a literal) you should not use it.

So, it's generally dangerous -and conceptually wrong- to pass a dynamic string as single argument of printf:

  char * myMessage;
  // ... myMessage gets assigned at runtime, unpredictable content
  printf(myMessage);  // <--- WRONG! (what if myMessage contains a '%' char?) 
  puts(myMessage);    // ok
  printf("%s\n",myMessage); // ok, equivalent to the previous, perhaps less efficient

The same applies to fputs vs fprintf (but fputs doesn't add the newline).

Besides formatting, puts returns a nonnegative integer if successful or EOF if unsuccessful; while printf returns the number of characters printed (not including the trailing null).

In simple cases, the compiler converts calls to printf() to calls to puts().

For example, the following code will be compiled to the assembly code I show next.

#include <stdio.h>
main() {
    printf("Hello world!");
    return 0;
}

push rbp
mov rbp,rsp
mov edi,str.Helloworld!
call dword imp.puts
mov eax,0x0
pop rbp
ret

In this example, I used GCC version 4.7.2 and compiled the source with gcc -o hello hello.c.